Chapter2:Kinematics

Kinematics…

• Branchofphysicsthatdealswithdescribingthemotionofobjects

• Motioncanbedescribedin3ways:a) Words/sentencesb) Mathematicalequationsc) graphically

Distance

• Theseparationbetween2points

• Expressesaquantityofmeasure(scalarquantity)

• Distancestraveledin1direction

Displacement• Theseparationbetweenanobjectandareferencepoint.

• Indicatesdistanceanddirection(vectorquantity)

• Usedwhendistancetraveledisnotequaltodistancefromstart

• Referencepoint– zeropointusedtodescribemotioninaframeofreference

∆d=df – di

∆d=displacement(m)df =finalpositiondi =initialposition

3m(N)

7 m (E)

d =7.6m(NE)

di

df

Pg.40PracticeProblems2.1.1

Speed• Thedistanceanobjecttravelsinaunitoftime

• Instantaneousspeedishowfastyourgoingataninstantoftime(i.e.,speedometerreadingincar)

• Average speedisthetotaldistancetraveledoverthetotaltime.

v = ∆d∆t

v=averagevelocity(m/s)∆d =displacement(m)∆t =timetakenforchange(s)

Example:Acartravels25kmin0.70h,thentravels35kminthenext1.50h.Whatistheaveragevelocityoftheentiretrip?

Example:Howfarwillawomantravelin15minutesifsheisdrivinghercardownthehighwayat24m/s?

Example:Howlongdoesittakeagirltotravel1800m,ifsheisridingherbikeatarateof2.50m/s?

Velocity

• indicatesspeed anddirection ofamovingobject.

• hasmagnitudeanddirection(vectorquantity)

• equaltotheslope ofadisplacement-timegraph.

ConstantVelocity• Achievedwhentheaveragevelocityofanobjectisthesameforallintervals(uniformmotion)

• Constantvelocitiesproducestraightlinesondisplacementtimegraphs.

d

t

• Steeperslope= greatervelocity

• Horizontalslope= zerovelocity(standingstill)

• Positivevelocity= indicatesmotionawayfromthestart

• Negativevelocity= indicatesmotiontowardsstartingpoint

VA =positivevelocity

VB =zerovelocity

VC =negativevelocity

VD =fastestvelocity

VE =slowest(non-zero)velocity t

d

A

B

C

D

E

SlopeofChangingVelocity

• Whenvelocitychangestheobjectisaccelerating

• Thelineonadisplacement-timegraphwillbecurved.

d

t

Tangent– InstantaneousVelocity

Tangent – straightlinethathasthesameslopeasapointonacurve

• Drawalinethroughasinglepoint,butdonotcrossthegraph– justtouchthepoint.

• Thevelocityatthatpoint isequaltotheslopeofthetangent.

d

t

tangent

VA =5.4m– 1.2m = 4.2m =0.66m/s10.9s– 4.5s 6.4s

VB =8.5m– 8.5m = 0m =0m/s17.0s– 7.0s 10s

VC =4.0m– 7.2m = -3.2m =-0.68m/s16.7s– 12.0s 4.7s

Acceleration

• therate atwhichvelocity changes(increasesordecreases).

• changingvelocity=acceleration

• constantvelocity=zeroacceleration

a= ∆vt

a=averageacceleration(m/s2)∆v = changeinvelocity(m/s)t=timetakenforchange(s)vf =finalvelocityvi =initialvelocity

=vf - vit

Example:Acaracceleratesfrom20km/hto80km/hin10seconds.Calculatetheaverageacceleration.(Assume2sig.figs.forallmeasurements)

Example2:Arunneracceleratesfrom0.52m/sto0.78m/s in0.050s.Whatisheracceleration?

Example3:Acaracceleratesfromrestatarateof50.0cm/s2 for12.5s.Howfastisitnowmoving?

Example4:Aturtlewantstoacceleratefrom2mm/s to8mm/s.Howlongdoesittake,ifitsmaximumaccelerationis3mm/s2?

AccelerationandSlope

• Accelerationisequaltotheslopeofavelocity-timegraph

v

t

A:constantaccelerationB:greaterconstantacceleration

v

t

A

B

PositiveAcceleration• Constantaccelerationproducesastraightline(increaseofvelocityisthesameforeachunitoftime).

• Changingacceleration producesacurvedline(increaseinvelocityisnotthesameforeachunit).

v v

tt

NegativeAcceleration• Slowingdownproducesnegativeacceleration

• Straightlinewhendecreaseinvelocityisthesameforeachunitoftime.

• Curvedlinewhendecreaseinvelocitychangeswithtime.

v v

tt

ZeroAcceleration

• Velocityremainsconstant.

v

t

ChangeinDirection

• Ifdirectionchanges,thevelocitychangesandtherefore,thereisacceleration.

i.e.,Ferriswheelrotatesataconstantspeedinacircularmotion,butthedirectionchanges,resultingin

acceleration.

ChangingAcceleration

• CurveonaV-Tgraph

• Averageacceleration=slopeofthestraightlinejoiningtwopointsonacurveofaV-Tgraph

v

t

ChangingAcceleration

• Instantaneousacceleration=slopeofthetangentofapointonthegraph.

v

t

2. Whatistheaccelerationforthefirst7.0s?(5.2- 0m/s)/7.0s

=0.74m/s2

3.Whatistheaverageaccelerationforeachofthefollowingtimeintervals?

a) 5.0to15.0s?b) 9.0to13.0s?c) 15.0to20.0s?

(7.8-3.9m/s)/10.0s=0.39m/s2

(5.0-6.3m/s)/4.0s=-0.33m/s2

(11.0-7.2m/s)/5.0s=0.76m/s2

4.Whatistheaccelerationforeachofthefollowingtimes?

a) 15.0s?b) 11.0s?c) 17.0s?

(10.0-3.0m/s)/(16.0-13.0s)=2.3m/s2

(3.0-8.8m/s)/(15.0-6.0s)=-0.64m/s2

VelocitywithUniformAcceleration

• Velocity-timegraphisastraightline• Slopeoftheline=acceleration

a= vf – vit

vf =vi +at

vf =finalvelocity(m/s)vi =initialvelocity(m/s)a=acceleration(m/s2)t=time(s)

Ex#1.Acarwithavelocityof5.0m/s,acceleratesatarateof5.0m/s2.Whatisthevelocityofthecaratatimeof3.5s?

Ex#2.Asupersonicjetthatisflyingat145m/sisaccelerateduniformlyattherateof23.1m/s2 for20.0s.Whatisitsfinalvelocity?

DisplacementDuringConstantAcceleration

Displacement:GivenVelocityandTime

• displacementisequaltothetotalarea underthelineofavelocity-timegraph.

• Duringconstantvelocityfromrest:

vave = dt

d=vave t

d=½(vf +vi)t

vf

vit

Ex#1.Howfardoesadragstertravelin6.00s,acceleratingsteadilyfromzeroto90.0m/s?

Ex#2.Twoskateboardersacceleratesteadilyfrom4.5m/sto11.5m/sin6.0s.Howfardotheytravel?

Displacement:GivenAccelerationandTime

If vf =vi +at

and d=½(vf+vi)t

then d=½(vi +at+vi)t

d=½(2vi +at)t

d=½(2vit +at2)

d=vit +½at2

Ex#1.Askieracceleratesat1.20m/s2downanicyslopestartingfromrest.Howfardoesshegetin5.0s?

Ex#2.Whatistheaccelerationofanobjectthatacceleratessteadilyfromrest,travelingadistanceof150mover10.0s?

Ex#3.Howlongdoesittakeanairplaneacceleratingfromrestat5.0m/s2 totravel300m?

Displacement:GivenVelocityandAcceleration

If vf =vi +at

t =v - via

and

d= vf - via

æ

è ç

ö

ø ÷ vi +

12a vf - vi

a

æ

è ç

ö

ø ÷

é

ë ê

ù

û ú

d= vit+ 12at2

= (t)(vi +12at)

ad= vf - vi( ) vi + 12vf - vi( )

é

ë ê

ù

û ú

ad= vf - vi( ) vi +12vf -

12vi

æ

è ç

ö

ø ÷

ad= vf - vi( ) 12vf + 1

2vi

æ

è ç

ö

ø ÷

ad= 12vf - vi( ) vf + vi( )

2ad= vf - vi( ) vf + vi( )

2ad= vf2- vi

2

vf

2= vi

2+ 2ad

Ex#1.Abulletacceleratesat6.8x104 m/s2 fromrestasittravelsthe0.80m oftheriflebarrel.Whatvelocitydoesthebullethaveasitleavesthebarrel?

Ex#2.Adrivertravelingat95km/hseesadeerstandingontheroad150mahead.Heslamsonhisbreaksanddeceleratesatarateof-2.0m/s2.Willhestopintime?

AccelerationduetoGravity

• Galileoshowedthatallobjectsfalltoearthwithaconstantacceleration,ifairresistancecanbeignored.

• g isthesymbolforaccelerationduetogravity.

• Onthesurfaceoftheearth,g =9.80m/s2

(variesslightly,dependingondistancefromcentreofearth)

Assumingnoairresistance,allaccelerationformulasapplytofallobjects(substituteg fora)

Ex#1:The“Hellevator”rideatPlayland fallsfreelyforatimeof1.8s.

a) Whatisthevelocityattheendofthedrop?

Ex#2:The“Hellevator”rideatPlayland fallsfreelyforatimeof1.8s.

b) Howfardiditfall?

Ex#2:Acoyotedropsananviloffacliffthatis150mhigh,hopingtohitaroadrunnerbelow.Howlongdoestheroadrunnerhavetogetoutofthewayifhedoesn’twanttogetcrushed?